A polygonal region is a plane figure that can be written as a union of a finitenumber of triangular regions, subject to a constraint. The constraint is that
the triangular regions are nonoverlapping. That means that if two triangular regions
intersect, then their intersection is either a vertex or edge of each of the triangular regions.

A given polygonal region can of course be written as such a union in infinitely many ways.

We wish to assign a number to each polygonal region that corresponds to our intuitive idea of
”area”. To do that we first write down a set of postulates that ”area” should satisfy and then
show that the postulates can be satisfied. We let ℛℛ\mathscr{R} denote the set of polygonal regions.

1.

There is a functionα:ℛ→ℝnormal-:αnormal-→ℛℝ\alpha:\mathscr{R}\to\mathbb{R}.

2.

α>0α0\alpha>0.

3.

If two triangular regions are congruent, then they have the same area.

4.

If two polygonal regions R1subscriptR1R_{1} and R2subscriptR2R_{2} are nonoverlapping then
α⁢(R1∪R2)=α⁢(R1)+α⁢(R2)αsubscriptR1subscriptR2αsubscriptR1αsubscriptR2\alpha(R_{1}\cup R_{2})=\alpha(R_{1})+\alpha(R_{2}).

The fifth postulate is needed because if αα\alpha satisfies the first 4 postulates, then so does
2⁢α2α2\alpha. So, this postulate serves to make αα\alpha a unique function.

Theorem 1.

If a square has sides of length sss then its area is s2superscripts2s^{2}.

Proof. We first show that if a square has sides of length 1q1q\frac{1}{q} then its area is
1q21superscriptq2\frac{1}{q^{2}}. A unit square can be decomposed into q2superscriptq2q^{2} squares, each having side of length
1q1q\frac{1}{q}. Each of the smaller squares has the same area, AAA, by postulates 3 and 4.
Hence,

1=q2⁢A,1superscriptq2A1=q^{2}A,

so that A=1q2A1superscriptq2A=\frac{1}{q^{2}}. Next, if a square has sides of length pqpq\frac{p}{q} then its area
is p2q2superscriptp2superscriptq2\frac{p^{2}}{q^{2}}. This is because the square can be written as p2superscriptp2p^{2} squares each having a side of length
1q1q\frac{1}{q}. If AAA is the area then

Finally, let TTT be a square of side sss and Tp/qsubscriptTpqT_{{p/q}} denote a square of side pqpq\frac{p}{q} having
one angle in common with TTT. In the following sequence of statements, each statement
is equivalent to the next one:

1.

pq<s.pqs\frac{p}{q}<s.

2.

Tp/q⊂T.subscriptTpqTT_{{p/q}}\subset T.

3.

α⁢(Tp/q)<α⁢(T).αsubscriptTpqαT\alpha(T_{{p/q}})<\alpha(T).

4.

p2q2<α⁢(T).superscriptp2superscriptq2αT\frac{p^{2}}{q^{2}}<\alpha(T).

5.

pq<α⁢(T).pqαT\frac{p}{q}<\sqrt{\alpha(T)}.

Hence, statements 1 and 5 are equivalent so that

s=α⁢(T),sαTs=\sqrt{\alpha(T)},

so that s2=α⁢(T)superscripts2αTs^{2}=\alpha(T).
This proves the theorem.

One can then proceed on the basis of the postulates to show the following sequence of results.

Theorem 2.

The area of a rectangular region is the product of its base and its altitude.

Rather than speak of the area of a triangular region or the area of a rectangle region,
we shall refer to the area of a triangle, or area of a rectangle, or other polygonal
region, as is usually done.

Theorem 3.

The area of a right triangle is 1/2 product of the lengths of its legs.

Theorem 4.

The area of a triangle is 1/2 the product of any base and the corresponding altitude.

Theorem 5.

The area of a parallelogram is the product of a base and the corresponding altitude.

Theorem 6.

The area of a trapezoid is 1/2 the product of the altitude and the sum of the bases.

Theorem 7.

If two triangles are similar then the ratio of their areas is the square of the ratio
of any two corresponding sides.

What remains to do is to show that there is a function αα\alpha that satisfies postulates
1 to 5.
It might seem reasonable to just define the area of a rectangle, but rectangles are not
useful for dissecting polygonal regions. The easier way to use triangles.
So one defines the area of a triangle to be
1/2 the product of the base and its corresponding altitude. For this to be well-defined
one has to show first that the product of a base and altitude (for a given triangle)
does not depend on which base is chosen. Next, one defines the area of a polygonal region
as the sum of the areas of the triangular regions in some decomposition. Of course one has to
show that this sum does not depend on the particular decomposition that is used.
Given this definition for αα\alpha, it is then a simple matter to show that
postulates 1 to 5 are satisfied.